# §10.49(i) Unmodified Functions

Define $a_{k}(\nu)$ as in (10.17.1). Then

 10.49.1 $a_{k}(n+\tfrac{1}{2})=\begin{cases}\dfrac{(n+k)!}{2^{k}k!(n-k)!},&k=0,1,\ldots% ,n,\\ 0,&k=n+1,n+2,\ldots.\end{cases}$
 10.49.2 $\mathop{\mathsf{j}_{n}\/}\nolimits\!\left(z\right)=\mathop{\sin\/}\nolimits\!% \left(z-\tfrac{1}{2}n\pi\right)\sum_{k=0}^{\left\lfloor n/2\right\rfloor}(-1)^% {k}\frac{a_{2k}(n+\tfrac{1}{2})}{z^{2k+1}}+\mathop{\cos\/}\nolimits\!\left(z-% \tfrac{1}{2}n\pi\right)\sum_{k=0}^{\left\lfloor(n-1)/2\right\rfloor}(-1)^{k}% \frac{a_{2k+1}(n+\tfrac{1}{2})}{z^{2k+2}}.$
 10.49.3 $\displaystyle\mathop{\mathsf{j}_{0}\/}\nolimits\!\left(z\right)$ $\displaystyle=\frac{\mathop{\sin\/}\nolimits z}{z},$ $\displaystyle\mathop{\mathsf{j}_{1}\/}\nolimits\!\left(z\right)$ $\displaystyle=\frac{\mathop{\sin\/}\nolimits z}{z^{2}}-\frac{\mathop{\cos\/}% \nolimits z}{z},$ $\displaystyle\mathop{\mathsf{j}_{2}\/}\nolimits\!\left(z\right)$ $\displaystyle=\left(-\frac{1}{z}+\frac{3}{z^{3}}\right)\mathop{\sin\/}% \nolimits z-\frac{3}{z^{2}}\mathop{\cos\/}\nolimits z.$ Symbols: $\mathop{\cos\/}\nolimits z$: cosine function, $\mathop{\sin\/}\nolimits z$: sine function, $\mathop{\mathsf{j}_{n}\/}\nolimits\!\left(z\right)$: spherical Bessel function of the first kind and $z$: complex variable A&S Ref: 10.1.11 (corrected) Referenced by: §10.49(iii), §10.56 Permalink: http://dlmf.nist.gov/10.49.E3 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png
 10.49.4 $\mathop{\mathsf{y}_{n}\/}\nolimits\!\left(z\right)=-\mathop{\cos\/}\nolimits\!% \left(z-\tfrac{1}{2}n\pi\right)\sum_{k=0}^{\left\lfloor n/2\right\rfloor}(-1)^% {k}\frac{a_{2k}(n+\tfrac{1}{2})}{z^{2k+1}}+\mathop{\sin\/}\nolimits\!\left(z-% \tfrac{1}{2}n\pi\right)\sum_{k=0}^{\left\lfloor(n-1)/2\right\rfloor}(-1)^{k}% \frac{a_{2k+1}(n+\tfrac{1}{2})}{z^{2k+2}}.$
 10.49.5 $\displaystyle\mathop{\mathsf{y}_{0}\/}\nolimits\!\left(z\right)$ $\displaystyle=-\frac{\mathop{\cos\/}\nolimits z}{z},$ $\displaystyle\mathop{\mathsf{y}_{1}\/}\nolimits\!\left(z\right)$ $\displaystyle=-\frac{\mathop{\cos\/}\nolimits z}{z^{2}}-\frac{\mathop{\sin\/}% \nolimits z}{z},$ $\displaystyle\mathop{\mathsf{y}_{2}\/}\nolimits\!\left(z\right)$ $\displaystyle=\left(\frac{1}{z}-\frac{3}{z^{3}}\right)\mathop{\cos\/}\nolimits z% -\frac{3}{z^{2}}\mathop{\sin\/}\nolimits z.$ Symbols: $\mathop{\cos\/}\nolimits z$: cosine function, $\mathop{\sin\/}\nolimits z$: sine function, $\mathop{\mathsf{y}_{n}\/}\nolimits\!\left(z\right)$: spherical Bessel function of the second kind and $z$: complex variable A&S Ref: 10.1.12 (corrected) Referenced by: §10.50, §10.56 Permalink: http://dlmf.nist.gov/10.49.E5 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png
 10.49.6 $\displaystyle\mathop{{\mathsf{h}^{(1)}_{n}}\/}\nolimits\!\left(z\right)$ $\displaystyle=e^{iz}\sum_{k=0}^{n}i^{k-n-1}\frac{a_{k}(n+\frac{1}{2})}{z^{k+1}},$ 10.49.7 $\displaystyle\mathop{{\mathsf{h}^{(2)}_{n}}\/}\nolimits\!\left(z\right)$ $\displaystyle=e^{-iz}\sum_{k=0}^{n}(-i)^{k-n-1}\frac{a_{k}(n+\frac{1}{2})}{z^{% k+1}}.$

# §10.49(ii) Modified Functions

Again, with $a_{k}(n+\tfrac{1}{2})$ as in (10.49.1),

 10.49.8 $\mathop{{\mathsf{i}^{(1)}_{n}}\/}\nolimits\!\left(z\right)=\tfrac{1}{2}e^{z}% \sum_{k=0}^{n}(-1)^{k}\frac{a_{k}(n+\frac{1}{2})}{z^{k+1}}+(-1)^{n+1}\*\tfrac{% 1}{2}e^{-z}\sum_{k=0}^{n}\frac{a_{k}(n+\frac{1}{2})}{z^{k+1}}.$
 10.49.9 $\displaystyle\mathop{{\mathsf{i}^{(1)}_{0}}\/}\nolimits\!\left(z\right)$ $\displaystyle=\frac{\mathop{\sinh\/}\nolimits z}{z},$ $\displaystyle\mathop{{\mathsf{i}^{(1)}_{1}}\/}\nolimits\!\left(z\right)$ $\displaystyle=-\frac{\mathop{\sinh\/}\nolimits z}{z^{2}}+\frac{\mathop{\cosh\/% }\nolimits z}{z},$ $\displaystyle\mathop{{\mathsf{i}^{(1)}_{2}}\/}\nolimits\!\left(z\right)$ $\displaystyle=\left(\frac{1}{z}+\frac{3}{z^{3}}\right)\mathop{\sinh\/}% \nolimits z-\frac{3}{z^{2}}\mathop{\cosh\/}\nolimits z.$
 10.49.10 $\mathop{{\mathsf{i}^{(2)}_{n}}\/}\nolimits\!\left(z\right)=\tfrac{1}{2}e^{z}% \sum_{k=0}^{n}(-1)^{k}\frac{a_{k}(n+\frac{1}{2})}{z^{k+1}}+(-1)^{n}\tfrac{1}{2% }e^{-z}\sum_{k=0}^{n}\frac{a_{k}(n+\frac{1}{2})}{z^{k+1}}.$
 10.49.11 $\displaystyle\mathop{{\mathsf{i}^{(2)}_{0}}\/}\nolimits\!\left(z\right)$ $\displaystyle=\frac{\mathop{\cosh\/}\nolimits z}{z},$ $\displaystyle\mathop{{\mathsf{i}^{(2)}_{1}}\/}\nolimits\!\left(z\right)$ $\displaystyle=-\frac{\mathop{\cosh\/}\nolimits z}{z^{2}}+\frac{\mathop{\sinh\/% }\nolimits z}{z},$ $\displaystyle\mathop{{\mathsf{i}^{(2)}_{2}}\/}\nolimits\!\left(z\right)$ $\displaystyle=\left(\frac{1}{z}+\frac{3}{z^{3}}\right)\mathop{\cosh\/}% \nolimits z-\frac{3}{z^{2}}\mathop{\sinh\/}\nolimits z.$
 10.49.12 $\mathop{\mathsf{k}_{n}\/}\nolimits\!\left(z\right)=\tfrac{1}{2}\pi e^{-z}\sum_% {k=0}^{n}\frac{a_{k}(n+\frac{1}{2})}{z^{k+1}}.$
 10.49.13 $\displaystyle\mathop{\mathsf{k}_{0}\/}\nolimits\!\left(z\right)$ $\displaystyle=\tfrac{1}{2}\pi\frac{e^{-z}}{z},$ $\displaystyle\mathop{\mathsf{k}_{1}\/}\nolimits\!\left(z\right)$ $\displaystyle=\tfrac{1}{2}\pi e^{-z}\left(\frac{1}{z}+\frac{1}{z^{2}}\right),$ $\displaystyle\mathop{\mathsf{k}_{2}\/}\nolimits\!\left(z\right)$ $\displaystyle=\tfrac{1}{2}\pi e^{-z}\left(\frac{1}{z}+\frac{3}{z^{2}}+\frac{3}% {z^{3}}\right).$ Symbols: $e$: base of exponential function, $\mathop{\mathsf{k}_{n}\/}\nolimits\!\left(z\right)$: modified spherical Bessel function and $z$: complex variable A&S Ref: 10.2.17 Permalink: http://dlmf.nist.gov/10.49.E13 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png

$\sum_{k=0}^{n}a_{k}(n+\tfrac{1}{2})z^{n-k}$ is sometimes called the Bessel polynomial of degree $n$. For a survey of properties of these polynomials and their generalizations see Grosswald (1978). See also §18.34, de Bruin et al. (1981a, b), and Dunster (2001c).

# §10.49(iii) Rayleigh’s Formulas

 10.49.14 $\displaystyle\mathop{\mathsf{j}_{n}\/}\nolimits\!\left(z\right)$ $\displaystyle=z^{n}\left(-\frac{1}{z}\frac{d}{dz}\right)^{n}\frac{\mathop{\sin% \/}\nolimits z}{z},$ $\displaystyle\mathop{\mathsf{y}_{n}\/}\nolimits\!\left(z\right)$ $\displaystyle=-z^{n}\left(-\frac{1}{z}\frac{d}{dz}\right)^{n}\frac{\mathop{% \cos\/}\nolimits z}{z}.$
 10.49.15 $\displaystyle\mathop{{\mathsf{i}^{(1)}_{n}}\/}\nolimits\!\left(z\right)$ $\displaystyle=z^{n}\left(\frac{1}{z}\frac{d}{dz}\right)^{n}\frac{\mathop{\sinh% \/}\nolimits z}{z},$ $\displaystyle\mathop{{\mathsf{i}^{(2)}_{n}}\/}\nolimits\!\left(z\right)$ $\displaystyle=z^{n}\left(\frac{1}{z}\frac{d}{dz}\right)^{n}\frac{\mathop{\cosh% \/}\nolimits z}{z}.$
 10.49.16 $\mathop{\mathsf{k}_{n}\/}\nolimits\!\left(z\right)=(-1)^{n}\tfrac{1}{2}\pi z^{% n}\left(\frac{1}{z}\frac{d}{dz}\right)^{n}\frac{e^{-z}}{z}.$

# §10.49(iv) Sums or Differences of Squares

Denote

 10.49.17 $s_{k}(n+\tfrac{1}{2})=\frac{(2k)!(n+k)!}{2^{2k}(k!)^{2}(n-k)!},$ $k=0,1,\ldots,n$. Symbols: $!$: $n!$: factorial, $n$: integer, $k$: nonnegative integer and $s_{k}(n)$ Permalink: http://dlmf.nist.gov/10.49.E17 Encodings: TeX, pMML, png

Then

 10.49.18 ${\mathop{\mathsf{j}_{n}\/}\nolimits^{2}}\!\left(z\right)+{\mathop{\mathsf{y}_{% n}\/}\nolimits^{2}}\!\left(z\right)=\sum_{k=0}^{n}\frac{s_{k}(n+\frac{1}{2})}{% z^{2k+2}}.$
 10.49.19 $\displaystyle{\mathop{\mathsf{j}_{0}\/}\nolimits^{2}}\!\left(z\right)+{\mathop% {\mathsf{y}_{0}\/}\nolimits^{2}}\!\left(z\right)$ $\displaystyle=z^{-2},$ $\displaystyle{\mathop{\mathsf{j}_{1}\/}\nolimits^{2}}\!\left(z\right)+{\mathop% {\mathsf{y}_{1}\/}\nolimits^{2}}\!\left(z\right)$ $\displaystyle=z^{-2}+z^{-4},$ $\displaystyle{\mathop{\mathsf{j}_{2}\/}\nolimits^{2}}\!\left(z\right)+{\mathop% {\mathsf{y}_{2}\/}\nolimits^{2}}\!\left(z\right)$ $\displaystyle=z^{-2}+3z^{-4}+9z^{-6}.$
 10.49.20 $\left(\mathop{{\mathsf{i}^{(1)}_{n}}\/}\nolimits\!\left(z\right)\right)^{2}-% \left(\mathop{{\mathsf{i}^{(2)}_{n}}\/}\nolimits\!\left(z\right)\right)^{2}=(-% 1)^{n+1}\sum_{k=0}^{n}(-1)^{k}\frac{s_{k}(n+\frac{1}{2})}{z^{2k+2}}.$
 10.49.21 $\displaystyle\Big(\mathop{{\mathsf{i}^{(1)}_{0}}\/}\nolimits\!\left(z\right)% \Big)^{2}-\Big(\mathop{{\mathsf{i}^{(2)}_{0}}\/}\nolimits\!\left(z\right)\Big)% ^{2}$ $\displaystyle=-z^{-2},$ $\displaystyle\Big(\mathop{{\mathsf{i}^{(1)}_{1}}\/}\nolimits\!\left(z\right)% \Big)^{2}-\Big(\mathop{{\mathsf{i}^{(2)}_{1}}\/}\nolimits\!\left(z\right)\Big)% ^{2}$ $\displaystyle=z^{-2}-z^{-4},$ $\displaystyle\Big(\mathop{{\mathsf{i}^{(1)}_{2}}\/}\nolimits\!\left(z\right)% \Big)^{2}-\Big(\mathop{{\mathsf{i}^{(2)}_{2}}\/}\nolimits\!\left(z\right)\Big)% ^{2}$ $\displaystyle=-z^{-2}+3z^{-4}-9z^{-6}.$